Question: A construction worker tosses a brick from a tall building. The brick's height (in meters above the ground) $t$ seconds after being thrown is modeled by $h(t)=-5t^2+20t+105$ Suppose we want to know the height of the brick above the ground at its highest point. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. $h(t)=$ 2) At its highest point, how far above the ground was the brick?
Explanation: Choosing a form The brick's highest point relates to the maximum of the function. Which form reveals this feature? Here's a summary of what each form reveals along with examples. Note that these are all equivalent forms of the same function, but not the function modeling the height of the brick. Form Example Feature revealed Standard $f(x)=2x^2-12x+{10}$ $y$ -intercept is ${10}$ Factored $f(x)=2(x-C{1})(x-C{5})$ Zeros are $x=C1$ and $x=C5$ Vertex $f(x)=2(x-{3})^2{-8}$ Vertex is $(3,{-8})$ Rewrite in vertex form The vertex of the function tells us the value of $t$ where the function reaches its maximum height, so let's rewrite $h(t)$ in vertex form by completing the square. The number that will help us complete the square is $\left(\dfrac{{-4}}{2}\right)^2={4}$ : $\begin{aligned} h(t)&=-5t^2+20t+105 \\\\ &=-5(t^2-4t)+105&&\text{Factor } -5 \text{ from first two terms}. \\\\ &=-5(t^2{-4}t+{4})+105{+20}&&\text{Complete the square}. \\\\ &=-5(t-2)^2+125&&\text{Factor and simplify}. \end{aligned}$ [How do we know what to add to complete the square?] How high is the highest point? The vertex form of the function reveals its vertex, and we know this point is a maximum for $h(t)$ since the leading coefficient $-5$ is negative. In general, for any quadratic function written in vertex form $f(x)=a(x-{h})^2+{k}$, we can conclude that the vertex is the point $({h},{k})$. So for $h(t)=-5(t-{2})^2+{125}$, the vertex is $({2},{125})$, and we know the brick's highest point was ${125}$ meters above the ground. Answers 1) The vertex form of the function reveals the highest point: $h(t)=-5\left(t-2\right)^2+125$ 2) At its highest point, the brick was $125$ meters above the ground.